Question 1

1 Write the formulas for the following coordination compounds:

(i) tetraamminediaquacobalt(III) chloride
(ii) potassium tetracyanidonickelate(II)
(iii) tris(ethane-1,2-diamine) chromium(III) chloride
(iv) ammnebromidochloridonitrito-N-platinate(II)
(v) dichloridobis(ethane-1,2-diamine)platinum(IV) nitrate
(vi) iron(III) hexacyanidoferrate(II)

Accepted Answer

Solutions:
(i) tetraamminediaquacobalt(III) chloride: The complex has 4 ammine (NH3) and 2 aqua (H2O) ligands coordinated to Co(III). The formula is [Co(NH3)4(H2O)2]Cl3.

(ii) potassium tetracyanidonickelate(II): The complex ion is [Ni(CN)4]2- with Ni(II). Potassium is the counter ion. Formula: K2[Ni(CN)4].

(iii) tris(ethane-1,2-diamine) chromium(III) chloride: Ethane-1,2-diamine (en) is a bidentate ligand. Three en ligands coordinate to Cr(III). Formula: [Cr(en)3]Cl3.

(iv) ammnebromidochloridonitrito-N-platinate(II): Ligands are ammine (NH3), bromo (Br-), chloro (Cl-), nitrito-N (NO2-) coordinated to Pt(II). Formula: [Pt(NH3)(Br)(Cl)(NO2)]

(v) dichloridobis(ethane-1,2-diamine)platinum(IV) nitrate: Two ethane-1,2-diamine (en) ligands and two chloride ligands coordinated to Pt(IV). Nitrate as counter ion. Formula: [PtCl2(en)2](NO3)2.

(vi) iron(III) hexacyanidoferrate(II): Complex ion is hexacyanidoferrate(II) [Fe(CN)6]4- with Fe(II), and Fe(III) is the counter ion. Formula: Fe3[Fe(CN)6]2. — Step-by-step reasoning:
(i) Co(III) with 4 NH3 and 2 H2O ligands, charge balanced by 3 Cl- ions.
(ii) Ni(II) with 4 CN- ligands, charge 2-, balanced by 2 K+ ions.
(iii) Cr(III) with 3 bidentate en ligands, charge 3+, balanced by 3 Cl- ions.
(iv) Pt(II) with four different ligands: NH3, Br-, Cl-, NO2-.
(v) Pt(IV) with 2 en (bidentate) and 2 Cl- ligands, charge balanced by 2 NO3- ions.
(vi) Fe(III) ions as counter ions to hexacyanidoferrate(II) complex ion.

This matches the coordination chemistry nomenclature and charge balance.

Question 2

2 Write the IUPAC names of the following coordination compounds:

(i)  [Co(NH3)6]Cl3
(ii)  [Co(NH3)5Cl]Cl2
(iii)  K3[Fe(CN)6]
(iv)  K3[Fe(C2O4)3]
(v)  K2[PdCl4]
(vi)  [Pt(NH3)2Cl(NH2CH3)]Cl

Accepted Answer

Solutions:
(i) [Co(NH3)6]Cl3: Hexaamminecobalt(III) chloride

(ii) [Co(NH3)5Cl]Cl2: Pentaamminechloridocobalt(III) chloride

(iii) K3[Fe(CN)6]: Potassium hexacyanoferrate(III)

(iv) K3[Fe(C2O4)3]: Potassium tris(oxalato)ferrate(III)

(v) K2[PdCl4]: Potassium tetrachloridopalladate(II)

(vi) [Pt(NH3)2Cl(NH2CH3)]Cl: Diamminemethylaminoplatinum(II) chloride — Step-by-step naming:
(i) Six ammine ligands on Co(III), chloride as counter ion.
(ii) Five ammine and one chloro ligand on Co(III), chloride as counter ion.
(iii) Hexacyanoferrate(III) complex ion with potassium counter ions.
(iv) Three oxalate ligands (bidentate) on Fe(III), potassium counter ions.
(v) Tetrachloridopalladate(II) complex ion with potassium counter ions.
(vi) Two ammine, one chloro, and one methylamine ligand on Pt(II), chloride counter ion.

This follows IUPAC nomenclature rules for coordination compounds.

Question 3

Indicate the types of isomerism exhibited by the following complexes and draw the structures for these isomers:

(i) K[Cr(H₂O)₂(C₂O₄)₂]
(ii) [Co(en)₃]Cl₃
(iii) [Co(NH₃)₅(NO₂)][(NO₃)₂]
(iv) [Pt(NH₃)(H₂O)Cl₂]

Accepted Answer

Solution:

(i) K[Cr(H₂O)₂(C₂O₄)₂]:
- Isomerism: This complex exhibits geometrical isomerism (cis-trans) because oxalate (C₂O₄) is a bidentate ligand and water is monodentate.
- Structures: Draw octahedral complexes with two water molecules adjacent (cis) and opposite (trans) to each other.

(ii) [Co(en)₃]Cl₃:
- Isomerism: Optical isomerism due to the presence of three bidentate ethylenediamine (en) ligands arranged in a chiral octahedral geometry.
- Structures: Show the Δ (right-handed) and Λ (left-handed) enantiomers.

(iii) [Co(NH₃)₅(NO₂)][(NO₃)₂]:
- Isomerism: Linkage isomerism because NO₂ can bind through N (nitro) or O (nitrito).
- Structures: Draw two isomers showing NO₂ bound via N and via O.

(iv) [Pt(NH₃)(H₂O)Cl₂]:
- Isomerism: Geometrical isomerism (cis-trans) in square planar complex.
- Structures: Draw cis and trans isomers with NH₃ and H₂O adjacent or opposite.

These isomers differ in spatial arrangement or ligand attachment, affecting properties. — The types of isomerism are identified based on ligand types and coordination geometry:
- Geometrical isomerism arises when ligands can be arranged differently around the central metal.
- Optical isomerism occurs when complexes are non-superimposable mirror images.
- Linkage isomerism occurs when ambidentate ligands bind through different atoms.
Drawing structures helps visualize these differences.

Question 4

Give evidence that [Co(NH₃)₅Cl]SO₄ and [Co(NH₃)₅(SO₄)]Cl are ionisation isomers.

Accepted Answer

Solution:

Ionisation isomers are compounds that have the same composition but differ in the exchange of ions inside and outside the coordination sphere.

Evidence:
- [Co(NH₃)₅Cl]SO₄ contains Cl⁻ inside the coordination sphere and SO₄²⁻ outside.
- [Co(NH₃)₅(SO₄)]Cl contains SO₄²⁻ inside the coordination sphere and Cl⁻ outside.

Tests:
- When dissolved in water, [Co(NH₃)₅Cl]SO₄ will give a positive test for SO₄²⁻ ions (e.g., with BaCl₂ forming BaSO₄ precipitate) but not for Cl⁻ ions.
- Conversely, [Co(NH₃)₅(SO₄)]Cl will give a positive test for Cl⁻ ions (e.g., with AgNO₃ forming AgCl precipitate) but not for SO₄²⁻ ions.

Thus, the difference in ions released on dissolution confirms they are ionisation isomers. — Ionisation isomers differ in which ions are inside or outside the coordination sphere.
Testing the solution for ions released confirms the identity of the ions outside the coordination sphere.
This difference in ion release is the evidence for ionisation isomerism.

Question 5

Explain on the basis of valence bond theory that $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$ ion with square planar structure is diamagnetic and the $\left[\mathrm{NiCl}_{4}\right]^{2-}$ ion with tetrahedral geometry is paramagnetic.

Accepted Answer

In $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$, Ni is in +2 oxidation state with electronic configuration $[Ar] 3d^{8} 4s^{0} 4p^{0}$. The strong field ligand CN⁻ causes pairing of electrons in the 3d orbitals, leading to dsp² hybridization and square planar geometry. All electrons are paired, making the complex diamagnetic. In contrast, in $\left[\mathrm{NiCl}_{4}\right]^{2-}$, Cl⁻ is a weak field ligand and does not cause pairing of electrons. Ni²⁺ has 3d⁸ configuration with four unpaired electrons occupying the orbitals. The hybridization is sp³ leading to tetrahedral geometry and the complex is paramagnetic due to unpaired electrons. — The difference in magnetic behavior arises from the ligand field strength. CN⁻ is a strong field ligand causing pairing of electrons and square planar geometry (dsp²), resulting in diamagnetism. Cl⁻ is a weak field ligand, no pairing occurs, sp³ hybridization leads to tetrahedral geometry and paramagnetism.

Question 6

$\left[\mathrm{NiCl}_{4}\right]^{2-}$ is paramagnetic while $\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]$ is diamagnetic though both are tetrahedral. Why?

Accepted Answer

Both complexes have Ni in zero or +2 oxidation state with tetrahedral geometry. However, CO is a strong field ligand causing pairing of electrons in Ni's d orbitals, resulting in diamagnetism. Cl⁻ is a weak field ligand and does not cause pairing, so $\left[\mathrm{NiCl}_{4}\right]^{2-}$ remains paramagnetic with unpaired electrons. The difference in ligand field strength explains the difference in magnetic properties despite similar geometry. — CO ligand causes pairing due to strong ligand field (π-acceptor), leading to diamagnetic complex. Cl⁻ ligand is weak field, no pairing, paramagnetic complex.

Question 7

$\left[\mathrm{Fe}(\mathrm{H}_{2}\mathrm{O})_{6}\right]^{3+}$ is strongly paramagnetic whereas $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$ is weakly paramagnetic. Explain.

Accepted Answer

In $\left[\mathrm{Fe}(\mathrm{H}_{2}\mathrm{O})_{6}\right]^{3+}$, Fe³⁺ has 3d⁵ configuration. Water is a weak field ligand, so electrons remain unpaired leading to five unpaired electrons and strong paramagnetism. In $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$, CN⁻ is a strong field ligand causing pairing of electrons, resulting in only one unpaired electron and weak paramagnetism. — The difference in magnetic behavior is due to ligand field strength: weak field H₂O ligand does not cause pairing; strong field CN⁻ ligand causes pairing of electrons.

Question 8

Explain $\left[\mathrm{Co}(\mathrm{NH}_{3})_{6}\right]^{3+}$ is an inner orbital complex whereas $\left[\mathrm{Ni}(\mathrm{NH}_{3})_{6}\right]^{2+}$ is an outer orbital complex.

Accepted Answer

In $\left[\mathrm{Co}(\mathrm{NH}_{3})_{6}\right]^{3+}$, Co³⁺ has a d⁶ configuration. NH₃ is a strong field ligand causing pairing of electrons and involvement of 3d orbitals (inner orbitals) in hybridization (d²sp³), making it an inner orbital complex. In $\left[\mathrm{Ni}(\mathrm{NH}_{3})_{6}\right]^{2+}$, Ni²⁺ has d⁸ configuration and NH₃ does not cause pairing of all electrons, so 4s and 4p orbitals (outer orbitals) are involved in hybridization (sp³d²), making it an outer orbital complex. — The difference arises due to oxidation state and ligand field strength affecting electron pairing and orbital involvement in hybridization.

Question 9

Predict the number of unpaired electrons in the square planar $\left[\mathrm{Pt}(\mathrm{CN})_{4}\right]^{2-}$ ion.

Accepted Answer

Pt in $\left[\mathrm{Pt}(\mathrm{CN})_{4}\right]^{2-}$ is in +2 oxidation state with electronic configuration 5d⁸. CN⁻ is a strong field ligand causing pairing of electrons. The complex has dsp² hybridization with square planar geometry. All electrons are paired, so the number of unpaired electrons is zero. — Strong field ligand CN⁻ causes pairing of electrons in Pt(II) d orbitals, resulting in diamagnetic square planar complex with no unpaired electrons.

Question 10

The hexaquo manganese(II) ion contains five unpaired electrons, while the hexacyano ion contains only one unpaired electron. Explain using Crystal Field Theory.

Accepted Answer

In the hexaquo manganese(II) ion $\left[\mathrm{Mn}(\mathrm{H}_{2}\mathrm{O})_{6}\right]^{2+}$, Mn²⁺ has 3d⁵ configuration. Water is a weak field ligand, so no pairing occurs and all five d electrons remain unpaired. In the hexacyano manganese(II) ion $\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{4-}$, CN⁻ is a strong field ligand causing pairing of four electrons, leaving only one unpaired electron. Crystal Field Theory explains this difference based on ligand field strength and splitting of d orbitals. — Weak field H₂O ligand leads to high spin complex with five unpaired electrons; strong field CN⁻ ligand leads to low spin complex with one unpaired electron due to electron pairing in lower energy orbitals.

Question 11

1 Explain the bonding in coordination compounds in terms of Werner's postulates.

Accepted Answer

Werner's theory explains the bonding in coordination compounds based on two main postulates:

1. Primary valency (oxidation state) is ionizable and corresponds to the charge on the metal ion.
2. Secondary valency corresponds to the coordination number and is satisfied by ligands bonded to the metal ion.

According to Werner, coordination compounds consist of a central metal atom/ion surrounded by ligands. The ligands are attached to the metal ion through coordinate covalent bonds (secondary valencies). The primary valencies are satisfied by ions that can be replaced by other ions in solution.

Thus, the bonding in coordination compounds involves coordinate bonds formed by the donation of lone pairs from ligands to the vacant orbitals of the metal ion, consistent with Werner's postulates. — Werner's postulates distinguish between primary valencies (oxidation state) and secondary valencies (coordination number). The primary valencies are ionizable and correspond to the charge on the metal ion, while the secondary valencies are satisfied by ligands attached via coordinate bonds. This explains the structure and bonding in coordination compounds.

Question 12

2 FeSO4 solution mixed with (NH4)2SO4 solution in 1:1 molar ratio gives the test of Fe2+ ion but CuSO4 solution mixed with aqueous ammonia in 1:4 molar ratio does not give the test of Cu2+ ion. Explain why?

Accepted Answer

When FeSO4 solution is mixed with (NH4)2SO4 in 1:1 molar ratio, Fe2+ ions remain free in solution and give the characteristic test for Fe2+ ion.

However, when CuSO4 solution is mixed with aqueous ammonia in 1:4 molar ratio, the Cu2+ ions form a complex ion [Cu(NH3)4]2+, which is stable and does not release free Cu2+ ions. Hence, the characteristic test for Cu2+ ion is not observed.

Thus, the difference arises because Fe2+ ions remain free while Cu2+ ions form a stable coordination complex with ammonia. — Fe2+ ions do not form a stable complex with sulfate ions and remain free, giving the test for Fe2+. Cu2+ ions form a stable tetraammine complex with ammonia, which prevents free Cu2+ ions from reacting in the test.

Question 13

3 Explain with two examples each of the following: coordination entity, ligand, coordination number, coordination polyhedron, homoleptic and heteroleptic.

Accepted Answer

Coordination entity: A species consisting of a central metal atom/ion bonded to ligands.
Examples: [Fe(CN)6]4-, [Co(NH3)6]3+

Ligand: An ion or molecule that donates a pair of electrons to the central metal atom/ion.
Examples: NH3, Cl-

Coordination number: Number of ligand donor atoms bonded to the central metal atom/ion.
Examples: 6 in [Fe(CN)6]4-, 4 in [Ni(CO)4]

Coordination polyhedron: The spatial arrangement of ligands around the central metal atom/ion.
Examples: Octahedral in [Co(NH3)6]3+, Tetrahedral in [Ni(CO)4]

Homoleptic complex: Complexes in which all ligands are identical.
Examples: [Fe(CN)6]4-, [Ni(CO)4]

Heteroleptic complex: Complexes containing more than one kind of ligand.
Examples: [Co(NH3)4Cl2]+, [Pt(NH3)2Cl2] — Each term is defined with examples to illustrate the concept clearly. Coordination entity refers to the entire complex ion or molecule, ligand is the donor species, coordination number is the count of donor atoms, coordination polyhedron is the geometry, homoleptic complexes have identical ligands, and heteroleptic complexes have different ligands.

Question 14

4 What is meant by unidentate, didentate and ambidentate ligands? Give two examples for each.

Accepted Answer

Unidentate ligands: Ligands that have only one donor atom to bind to the central metal atom/ion.
Examples: NH3, Cl-

Didentate ligands: Ligands that have two donor atoms which can bind simultaneously to the central metal atom/ion.
Examples: Ethylenediamine (en), Oxalate ion (C2O4)2-

Ambidentate ligands: Ligands that have more than one donor atom but can coordinate through only one donor atom at a time.
Examples: Thiocyanate ion (SCN-) which can bind through S or N, Nitrite ion (NO2-) which can bind through N or O. — Unidentate ligands bind through a single donor atom, didentate ligands have two donor atoms that bind simultaneously forming chelates, and ambidentate ligands can bind through different donor atoms but only one at a time.

Question 15

5 Specify the oxidation numbers of the metals in the following coordination entities:

(i) [Co(H2O)(CN)(en)2]2+

(ii) [CoBr2(en)2]+

(iii) [PtCl4]3-

(iv) K3[Fe(CN)6]

(v) [Cr(NH3)3Cl3]

Accepted Answer

(i) [Co(H2O)(CN)(en)2]2+:
H2O is neutral, CN- is -1, en (ethylenediamine) is neutral.
Let oxidation state of Co = x
x + 0 + (-1) + 0 = +2
x - 1 = 2
x = +3

(ii) [CoBr2(en)2]+:
Br- is -1 each, en is neutral.
Let oxidation state of Co = x
x + 2(-1) + 0 = +1
x - 2 = 1
x = +3

(iii) [PtCl4]3-:
Cl- is -1 each
Let oxidation state of Pt = x
x + 4(-1) = -3
x - 4 = -3
x = +1

(iv) K3[Fe(CN)6]:
K+ is +1 each, total +3
CN- is -1 each, total -6
Let oxidation state of Fe = x
3(+1) + x + 6(-1) = 0
3 + x - 6 = 0
x - 3 = 0
x = +3

(v) [Cr(NH3)3Cl3]:
NH3 is neutral, Cl- is -1 each
Let oxidation state of Cr = x
x + 0 + 3(-1) = 0
x - 3 = 0
x = +3 — Oxidation state is calculated by considering the charge on the complex and the charges of ligands. Neutral ligands contribute zero, anionic ligands contribute their charge, and the overall charge is used to solve for the metal oxidation state.

Werner’s Theory of Coordination Compounds

Werner’s Theory of Coordination Compounds — Study Notes

Coordination Compounds

Werner's Theory of Coordination Compounds

Definitions of Some Important Terms Pertaining to Coordination Compounds

Practice Questions — Werner’s Theory of Coordination Compounds

All 5 Chapters in Chemistry-I